Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 14


$y=-\displaystyle \frac{2}{5}x++1$

Work Step by Step

We solve the given equation for y (write it in slope-intercept form) $2x+5y=15$ $5y=-2x+15$ $y=-\displaystyle \frac{2}{5}x+3$ The slope of the line is $m=-\displaystyle \frac{2}{5}$, and any line parallel to it has the same slope. Now, find the wanted parallel line. The point-slope equation of a line containing the point $(x_{1},y_{1})$, with slope $m$ is $y=y_{1}+m(x-x_{1})$ Given $(x_{1},y_{1})=(5,-1)$ and $m=-\displaystyle \frac{2}{5}$ $y=-1+(-\displaystyle \frac{2}{5})(x-5)$ $y=-1-\displaystyle \frac{2}{5}x+2$ $y=-\displaystyle \frac{2}{5}x++1$
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